mass percent and formulas

CHEMISTRY 161 – Fall 2018 Dr. Mioy Huynh

Mass Percent and FormulasDR. MIOY T. HUYNHYALE UNIVERSITYCHEMISTRY 161

FALL 2018

www.mioy.org/chem161

http://www.mioy.org/chem161


Introduction to mass percent

Imagine a single molecule of methane: CH4

• Ask yourself: Does this molecule contain more hydrogen or more carbon?

• IT DEPENDS!• Technically, 4 out of 5 atoms are hydrogen (80%), but….


Introduction to mass percent

CHEMISTS CARE ABOUT MASS PERCENT!

%Mass = mass partmass whole×100%


How do I calculate the mass percentages for CH4?

• Remember that the molar mass of CH4 is 16.04 g/mol:1 mol CH4 = 1 mol C + 4 mol H

= 1 (12.01 g) + 4 (1.008 g) = 16.04 g


How do I calculate the mass percentages for CH4?

• Remember that the molar mass of CH4 is 16.04 g/mol:


1 mol CH4 = 1 mol C + 4 mol H= 1 (12.01 g) + 4 (1.008 g) = 16.04 g

%C → 1(12.01) g16.04 g ×100% = 74.90%C

%H → 4(1.008) g16.04 g ×100% = 25.10%H

100.0%total


How do I calculate the mass percentages for 2 mol CH4?



• The molar mass of CH4 is 16.04 g/mol, but now:2 mol CH4 = 2 mol C + 8 mol H

= 2 (12.01 g) + 8 (1.008 g) = 32.08 g



• The molar mass of CH4 is 16.04 g/mol, but now:



%C → 2(12.01) g32.08 g ×100% = 74.90%C

%H → 8(1.008) g32.08 g ×100% = 25.10%H

100.0%total



• The molar mass of CH4 is 16.04 g/mol, but now:



%C → 2(12.01) g32.08 g ×100% = 74.90%C

%H → 8(1.008) g32.08 g ×100% = 25.10%H

100.0%total

SAME THING!


TAKE-HOME MESSAGE

Percent composition is independent of the starting amount!

This is why we usually assume we have 100 g or 1 mol.These are just super easy numbers to work with.

Note: If you wanted to use a strange amount, like 0.27 mol or 74.5 g of substance, your answers would be the same but the math isn’t as convenient. BUT you’ll still be right. J


What is the mass percent of oxygen in each substance?H2O2 versus H2O



H2O2 (hydrogen peroxide)

The molar mass of H2O2 is 34.02 g/mol.

H2O (water)

The molar mass of H2O is 18.02 g/mol.





2(16.00) g34.02 g ×100% = 94.06% O

H2O (water)


16.00 g18.02 g×100% = 88.79% O





2(16.00) g34.02 g ×100% = 94.06% O

H2O (water)


16.00 g18.02 g×100% = 88.79% O

What is the mass percent of hydrogen in each substance?





2(16.00) g34.02 g ×100% = 94.06% O

2(1.008) g34.02 g ×100% = 5.93% H

H2O (water)


16.00 g18.02 g×100% = 88.79% O

2(1.008) g18.02 g ×100% = 11.19% H

What is the mass percent of hydrogen in each substance?

Note: The numbers will not always be exactly 100%.


Most often, we use mass percentages to help us figure out what compound we have.

These are called EMPIRICAL FORMULAS.


You have some “nitrogen oxide” compound and you want to figure out what it is (both formula and name).

You know that it’s 30.4% nitrogen by mass.



You know that it’s 30.4% nitrogen by mass.• We want to know: NxOy = formula? name?

• Remember: the amount doesn’t matter for percent composition!





• Let’s assume we have 100 g of our NxOy.

• This means that for every 100 g of NxOy, we have:

• 30.4 g of N

• 69.6 g of O





• Let’s assume we have 100 g of our NxOy.

• This means that for every 100 g of NxOy, we have:

• 30.4 g of N

• 69.6 g of O

Q: Is our formula then N30.4O69.6?

A: No! Why? A chemical formula represents number of atoms in a

compound, not the mass of each.

We must convert the masses to moles.




We must convert the masses to moles.N → 30.4 g N × 1 mol N

14.01 g N = 2.17mol N

O → 69.6 g O × 1 mol O16.00 g O = 4.35mol O





Q: So, is our formula then N2.17O4.35?A: No! Why? Atoms can’t be fractional.

We need a whole number ratio!

N → 30.4 g N × 1 mol N14.01 g N = 2.17mol N

O → 69.6 g O × 1 mol O16.00 g O = 4.35mol O




We need the simplest whole number ratio!




We need the simplest whole number ratio!N → 30.4 g N × 1 mol N

14.01 g N = 2.17 mol N → 2.17 mol N2.17 = 1N

O → 69.6 g O × 1 mol O16.00 g O = 4.35 mol O → 4.35 mol O

2.17 = 2O

Divide the number of moles by the SMALLEST value!




We need the simplest whole number ratio!

THIS IS IT!Our compound has the empirical formula: NO2 (nitrogen dioxide)

N → 30.4 g N × 1 mol N14.01 g N = 2.17 mol N → 2.17 mol N

2.17 = 1N


2.17 = 2O


Find the percent composition of dinitrogen tetroxide.

• The molar mass of dinitrogen tetroxide (N2O4) is 92.02 g/mol.




%N → 2(14.01) g92.02 g ×100% = 30.4%N

%O → 4(16.00) g92.02 g ×100% = 69.6%O

100.0%total




This is the same as NO2!Q: How do we differentiate between NO2 and N2O4?A: You can use the molar masses of NO2 and N2O4

%N → 2(14.01) g92.02 g ×100% = 30.4%N

%O → 4(16.00) g92.02 g ×100% = 69.6%O

100.0%total


Aluminum oxide (AlxOy) is 41.51% Al and 36.92% O.Determine the empirical formula.



General procedure:1. Assume a 100 g sample.

Al → 41.51 g Al × 1 mol Al26.98 g Al = 1.54 mol Al

O → 36.92 g O × 1 mol O16.00 g O = 2.31 mol O



General procedure:1. Assume a 100 g sample.2. Convert masses to moles.

Al → 41.51 g Al × 1 mol Al26.98 g Al = 1.54 mol Al

O → 36.92 g O × 1 mol O16.00 g O = 2.31 mol O



General procedure:1. Assume a 100 g sample.2. Convert masses to moles.3. Divide the mole amounts by the smallest mole value.

Al → 41.51 g Al × 1 mol Al26.98 g Al = 1.54 mol Al → 1.54 mol Al

1.54 = 1Al


1.54 = 1.5O



General procedure:1. Assume a 100 g sample.2. Convert masses to moles.3. Divide the mole amounts by the smallest mole value.4. Write empirical formula from simplest whole number ratio.


1.54 = 1Al


1.54 = 1.5O

Uh... these aren’t whole

numbers!





1.54 = 1Al ×2 → 2Al


1.54 = 1.5O ×2 → 3OMultiply to get

whole numbers!





1.54 = 1Al ×2 → 2Al


1.54 = 1.5O ×2 → 3OAl2O3


Calculate the empirical formula for cisplatin if it is found to be65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl by mass.


Calculate the empirical formula for cisplatin if it is found to be65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl by mass.

Assuming a 100 g sample of cisplatin (PtaNbHcCld):

Pt → 65.02 g Pt × 1 mol Pt195.1 g Pt = 0.3333 mol Pt → 0.3333 mol Pt

0.3333 = 1Pt

N → 9.34 g N × 1 mol N14.01 g N = 0.667 mol N → 0.667 mol N

0.3333 = 2N

H → 2.02 g H × 1 mol H1.008 g H = 2.00 mol H → 2.00 mol H

0.3333 = 6H

Cl → 23.63 g Cl × 1 mol Cl35.45 g Cl = 0.6666 mol Cl → 0.6666 mol Cl

0.3333 = 2Cl

PtN2H6Cl2


Calculate the empirical formula for a halohydrocarbon if it is71.65% Cl, 24.27% C, and 4.07% H by mass.


Calculate the empirical formula for a halohydrocarbon if it is71.65% Cl, 24.27% C, and 4.07% H by mass.

Assuming a 100 g sample of the halohydrocarbon (ClaCbHc):

The empirical formula is ClCH2.

Cl → 71.65 g Cl × 1 mol Cl35.45 g Cl = 2.021 mol Cl → 2.021 mol Cl

2.021 = 1Cl

C → 24.27 g C × 1 mol C12.01 g C = 2.021 mol C → 2.021 mol C

2.021 = 1C

H → 4.07 g H × 1 mol H1.008 g H = 4.04 mol H → 4.04 mol H

2.021 = 2H


Determine the molecular formula for the same halohydrocarbon ifit has a molar mass of 98.96 g/mol.



We determined previously that the empirical formula is ClCH2.The empirical formula mass is 49.48 g/mol.




The molecular formula is always a multiple of the empirical formula. So: (ClCH2)n





We can determine the multiple (n) by taking the ratio between the molecular formula’s molar mass and the empirical formula mass:

! = #$%&' (&))*(+,',-&% .$'(/%& (&))





We can determine the multiple (n) by taking the ratio between the molecular formula’s molar mass and the empirical formula mass:

! = #$%&' (&))*(+,',-&% .$'(/%& (&)) =

98.96 g49.48 g = 2

The molecular formula is (ClCH2)2 or Cl2C2H4.


For each of the following, the molecular formula is given.Determine the empirical formula for each compound.

H2O2

C6H6

C4H10

CCl4

C2H4Cl4O2



H2O2 34.02 g/mol

C6H6 78.11 g/mol

C4H10 58.12 g/mol

CCl4 153.81 g/mol

C2H4Cl4O2 201.85 g/mol



H2O2 34.02 g/mol HO 17.01 g/mol

C6H6 78.11 g/mol CH 13.02 g/mol

C4H10 58.12 g/mol C2H5 29.06 g/mol

CCl4 153.81 g/mol CCl4 153.81 g/mol

C2H4Cl4O2 201.85 g/mol CH2Cl2O 100.93 g/mol



H2O2 34.02 g/mol HO 17.01 g/mol n = 2

C6H6 78.11 g/mol CH 13.02 g/mol n = 6

C4H10 58.12 g/mol C2H5 29.06 g/mol n = 2

CCl4 153.81 g/mol CCl4 153.81 g/mol n = 1

C2H4Cl4O2 201.85 g/mol CH2Cl2O 100.93 g/mol n = 2

mass percent and formulas

Documents